1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

S9 6 ANSWERS

7c. z = ln2 + i (-~ + 2mr), where n is an integer.

9a. exp(z) = f ;h (zt = Jim ( f, ~ (zt) = lim

n = O k-oo n=O k-oo

k

"' L.J .l.zn n.!

n =O

(justify!) =

( k - oolim n = O f, ;hzn), because the conj ugate is a continuous function {ex plain).

T his last quantity, of course, equals exp (z).

00

z E fhzn oo

lla. Method 1: lim • - l = lim n = l = Jim E -\zn-l = 1. Justify the last

z- o z z- o z z -On=l n.

equality. Method 2: Using L'Hopital's rule (T heorem 3.2), lim e•-^1 =

z-o z

lim e' = 1.

z- o^1

13 c. (a+ ib) e<a+ib)z.

00 00

lS. E einz = E (ei•)n. This is a geometric series. Show that Im(z) > 0

n=O n=O

implies lei' I = lei(x+iy)I < 1, so that the series converges by Theorem 4. 1 2.

2 2

17. Show that e" -y sin2xy is the imaginary part of exp (z^2 ), and therefore

harmonic by Theorem 3.8.

Section 5.2. The C omple x Logarithm : page 168

l a. 2 +iI.

le. ln2+i3;.

le. ln3 + i (1+2n) n, where n is an integer.

lg. ln 4 + i (! + 2n) 11", where n is an integer.

3a. ( ~) (1 - i).

3 c. 1 + i (-! + 2n) 11", where n is an integer.

Sa. Since Arg{l+ i) = ~.the funct ion f(z) = log7(z- l - i) defined for

z = re^18 f= 1 + i, where ~ < (} <^9 ;, is analytic, and f' (z) = z-~-i for

those values of z.

Sc. We set f (z) = zLog (z), and deduce that f' (z) = 1 + Log (z) for z = re^18 f=

0, where -71" < (} < n.